If $f(x)=\sqrt{x-3}$, what is the smallest real number $x$ in the domain of $f(f(x))$?
Solution: Note that $f(x)$ is defined if and only if $x\ge 3$.

Thus $f(f(x)) = f(\sqrt{x-3})$ is defined if and only if $$\sqrt{x-3}\ge 3.$$ This is true if and only if $$x-3\ge 3^2,$$ or equivalently, if $x\ge 12$. So the smallest real number for which $f(f(x))$ is defined is $\boxed{12}$.